Mixed boundary condition and uniqueness of solution

\bf{Problem: } Let L=a^{ij}D_{ij}u+b^i(x)D_iu+c(x)u is uniformly elliptic in a bounded domain \Omega, c\leq 0, c/\lambda is bounded and Lu=0 in \Omega
(1) Let \partial \Omega=S_1\cup S_2 (S_1 non-empty) and assume an interior sphere condition at each point of S_2. Suppose u\in C^2(\Omega)\cap C^1(\Omega\cup S_2)\cap C^0(\overline{\Omega}) satisfies the mixed boundary condition
\displaystyle u=0 on S_1, \displaystyle \sum \beta_iD_i u=0 on S_2
where the vector \boldsymbol{\beta}(x)=(\beta_1(x),\cdots,\beta_n(x)) has a non-zero normal component(to the interior sphere) at each point x\in S_2, then u\equiv 0.

(2) Let \partial \Omega satisfy an interior sphere condition, and assume that u\in C^2(\Omega)\cap C^1(\overline{\Omega}) satisfies the regular oblique derivative boundary condition
\displaystyle \alpha(x)u+\sum \beta(x)D_iu=0 on \partial \omega
where \alpha(\vec{\beta}\cdot \vec{\nu}>;;;;;;;;;0, \nu is the outer normal vector. Then u\equiv 0.

\mathbf{Proof: } (1) Suppose u\not\equiv constant. If there exists x\in \Omega such that u(x)>;;;;;;;;0, then \exists\, x_0\in\partial S_2 such that \sup _{x\in \Omega}u=u(x_0). WLOG, assume x_0 is the origin and \vec{\nu} is pointing to the negative x_n-axis. And B is the interior ball at x_0\in S_2. Since u is not constant, we know that u(x)0. This means D_nu(x-0)>;0
Since u\in C^1(\Omega\cup S_2), then u\in C^1(B\cup x_0). So \displaystyle D_iu(x_0)=0 for i=1,2,\cdots,n-1. So \vec{beta}\cdot \vec{Du}=0 means that \beta_n=0, this contradicts to the fact that \vec{\beta} has non-zero component on \vec{nu}.
So u\leq 0 in \Omega. Similarly, -u\leq 0. So u\equiv 0.
(2) Use the same technique.

\bf{Remark:} Gilbarg, Trudinger’ book. Chapter 3, exercise 3.1.

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